# Prove $\inf (A + B) = \inf A + \inf B$ [duplicate]

Prove:

If $A$ and $B$ are nonempty subsets of $\mathbb{R}$ and are both bounded above, then $$\operatorname{lub}(A+B)=\operatorname{lub}(A)+\operatorname{lub}(B).$$

I am thinking I need to utilize the fact that $A+B=\{a+b: a \in A, b \in B\}$, I'm just not sure how this gives a proof.

Thanks.

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## marked as duplicate by mrf, Davide Giraudo, Asaf Karagila, Paul, Hagen von EitzenFeb 4 '13 at 10:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You can find some good starting points on how to format mathematics on the site here and here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. –  Zev Chonoles Feb 4 '13 at 5:54
Thanks, my LaTeX knowledge is passable, but it definitely needs some work –  Peej Gerard Feb 4 '13 at 6:01
$\sup$ is more conventional, see wiki –  Frank Science Feb 4 '13 at 6:13
–  mrf Feb 4 '13 at 8:30

## 1 Answer

First we show that $\sup(A + B) \leq \sup(A) + \sup(B)$:

Let $\gamma$ in $A + B$ be arbitrary. We see that $$\gamma = a + b \quad \text{where} \quad a \in A \text{ and } b \in B$$ Now, we know that $$\sup(A) \geq a \quad \text{and} \quad \sup(B) \geq b$$ so it follows that $$\gamma = a + b \leq \sup(A) + \sup(B)$$ Since $\gamma$ was arbitrary, this holds for all elements of $A+B$. We must have $$\sup(A + B) \leq \sup(A) + \sup(B)$$

Next, we show that $\sup(A + B) \geq \sup(A) + \sup(B)$:

Well, assume the contrary. Then we have $$\sup(A + B) < \sup(A) + \sup(B)$$ This implies that for all $\gamma \in A + B$ there exists some $a \in A$ and some $b \in B$ such that $\gamma < a + b$. But $(A + B) \ni \sigma = a + b$, so this is a contradiction.

Putting these two together, we see that $\sup(A + B) = \sup(A) + \sup(B)$ as we wanted.

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